Expansion — Set 4

Correct Answer: D. 1×10^-3

• **1×10⁻³** = Fractional change in length = αΔT = 2×10⁻⁵ × 50 = 10⁻³ = 1×10⁻³. • **αΔT = 2×10⁻⁵ × 50 = 1×10⁻³** — This means a 0.1% increase in length for every 50 K rise. • 💡 Wrong-option analysis: 1×10⁻⁵: that would be just α, without the 50 K factor; 1×10⁻⁴: off by a factor of 10 from the correct calculation; 1×10⁻²: 10 times too large — would require ΔT = 500 K.

Correct Answer: D. 1.2 mm

• **1.2 mm** = ΔL = αLΔT = 12×10⁻⁶ × 1.0 × 100 = 12×10⁻⁴ m = 1.2 mm. • **ΔL = 12×10⁻⁶ × 1.0 × 100 = 1.2×10⁻³ m = 1.2 mm** — The small fractional expansion of 0.12% corresponds to a noticeable 1.2 mm change in a 1 m rod. • 💡 Wrong-option analysis: 0.012 mm: 100 times too small — this would be αΔT only divided by 100; 0.12 mm: 10 times too small — an arithmetic error; 12 mm: 10 times too large — forgetting to account for the 10⁻⁶ factor correctly.

Correct Answer: D. ΔA = 2αAΔT

• **ΔA = 2αAΔT** = Area expansion in two perpendicular directions gives ΔA/A ≈ 2αΔT for small changes, so ΔA = 2αAΔT. • **β = 2α** — The superficial expansion coefficient β = 2α; substituting: ΔA = βAΔT = 2αAΔT. • 💡 Wrong-option analysis: ΔA = αΔT/A: wrong dimensions — dividing by A gives wrong units; ΔA = 3αAΔT: that is the volume expansion formula (ΔV = γVΔT = 3αVΔT); ΔA = αA/ΔT: dividing by ΔT inverts the temperature dependence.

Correct Answer: B. ΔV = γVΔT

• **ΔV = γVΔT** = Volume expansion is proportional to original volume V, cubical expansion coefficient γ, and temperature rise ΔT. • **γ ≈ 3α** — Substituting: ΔV = 3αVΔT; for a 1 m³ solid with α = 12×10⁻⁶ K⁻¹ heated 100 K, ΔV = 3.6×10⁻³ m³. • 💡 Wrong-option analysis: ΔV = γV/ΔT: dividing by ΔT inverts the dependence; ΔV = γΔT/V: dividing by V gives wrong dimensions; ΔV = V/(γΔT): this formula inverts everything.

Correct Answer: C. Less than real expansion

• **Less than real expansion** = Apparent expansion = Real expansion − Container expansion; since the container also expands (taking up some volume), the observed rise of liquid is less than the true expansion. • **γ_apparent = γ_liquid − γ_container** — For mercury in glass, γ_apparent ≈ 1.8×10⁻⁴ K⁻¹ (real ≈ 1.82×10⁻⁴ minus glass ≈ 0.025×10⁻⁴). • 💡 Wrong-option analysis: Greater than real expansion: apparent cannot be greater than real if the container expands; Independent of container material: it directly depends on the container's expansion coefficient; Equal to real expansion always: only if the container were perfectly rigid (zero expansion).

Correct Answer: C. Bending due to different expansions of two metals

• **Bending due to different expansions of two metals** = A bimetallic thermometer uses a strip of two bonded metals with different α values; when temperature changes, the unequal expansion causes bending that is calibrated to read temperature. • **Differential expansion** — The degree of bending is proportional to ΔT and the difference in expansion coefficients (α₁ − α₂); larger differences give more sensitive thermometers. • 💡 Wrong-option analysis: Change of electrical resistance: that describes a platinum resistance thermometer; Change of color with temperature: that describes liquid crystal thermometers; Evaporation of a liquid: evaporation is unrelated to bimetallic strip operation.

Correct Answer: A. ΔL/L = αΔT

• **ΔL/L = αΔT** = Thermal strain is the fractional change in length; for a freely expanding rod, ΔL = αLΔT, so ΔL/L = αΔT — the strain depends only on α and the temperature change. • **Thermal strain vs. thermal stress** — If free expansion is allowed, strain = αΔT and stress = 0; if expansion is prevented, strain = 0 but stress = YαΔT. • 💡 Wrong-option analysis: ΔL/L = αLΔT: multiplying by L gives the extension ΔL, not the strain ΔL/L; ΔL/L = α/(ΔT): dividing by ΔT inverts the dependence; ΔL/L = ΔT/α: this inverts the relationship between α and ΔT.

Correct Answer: D. 1/273 per °C near 0°C

• **1/273 per °C near 0°C** = From the ideal gas law (V = nRT/P), at constant P: dV/V = dT/T. Near 0°C, T ≈ 273 K, so γ_gas ≈ 1/273 per °C. • **γ_gas = 1/T** — At higher temperatures, γ decreases (e.g., at 300 K, γ ≈ 1/300 per K). This is far larger than solids' typical γ ≈ 3×10⁻⁵ K⁻¹. • 💡 Wrong-option analysis: 0 per °C: zero expansion would violate the ideal gas law; 1/100 per °C: would correspond to T = 100 K (−173°C); 273 per °C: that is the temperature in kelvin at 0°C, not the expansion coefficient.

Correct Answer: B. Expands on freezing

• **Expands on freezing** = Water is unusual — it expands (by about 9%) when it freezes into ice; in a closed pipe, this expansion creates enormous pressure, which can burst the pipe. • **9% volume increase** — This is why ice floats on water (ice is less dense), and why water pipes must be protected from freezing in cold climates. • 💡 Wrong-option analysis: Turns into steam inside pipes: steam forms at 100°C; pipes burst at 0°C when water freezes; Contracts on freezing: most substances contract on solidifying, but water is an exception; Stops exerting pressure when cold: cold water still exerts pressure, and freezing water exerts even greater pressure.

Correct Answer: C. Mercury expands more than the glass container

• **Mercury expands more than the glass container** = Both mercury and glass expand when heated, but mercury's cubical expansion coefficient (γ ≈ 1.82×10⁻⁴ K⁻¹) is much larger than glass's (γ ≈ 0.025×10⁻⁴ K⁻¹); the net apparent expansion drives mercury up the capillary. • **Apparent expansion = γ_mercury − γ_glass ≈ 1.8×10⁻⁴ K⁻¹** — This difference is large and nearly constant, making mercury thermometers accurate and reliable. • 💡 Wrong-option analysis: Mercury changes color when heated: mercury does not change color; Mercury does not expand at all: mercury expands significantly; Glass expands more than mercury: glass expands far less than mercury.